The probability of selecting integers $$a \in [-5, 30]$$ such that $$x^2 + 2(a+4)x - 5a + 64 > 0$$, for all $$x \in R$$, is:

For the quadratic $$x^2 + 2(a+4)x - 5a + 64 > 0$$ to hold for all $$x \in \mathbb{R}$$, the discriminant must be strictly negative (since the leading coefficient is positive):

$$\Delta = [2(a+4)]^2 - 4 \cdot 1 \cdot (-5a + 64) < 0$$

$$4(a+4)^2 + 4(5a - 64) < 0$$

$$(a+4)^2 + 5a - 64 < 0$$

$$a^2 + 8a + 16 + 5a - 64 < 0$$

$$a^2 + 13a - 48 < 0$$

Factoring: $$(a + 16)(a - 3) < 0$$

This inequality holds for $$-16 < a < 3$$.

Now, $$a$$ is an integer selected from $$[-5, 30]$$. The integers in this range are: $$-5, -4, -3, \ldots, 30$$, giving a total of $$30 - (-5) + 1 = 36$$ integers.

The integers satisfying $$-16 < a < 3$$ within $$[-5, 30]$$ are: $$-5, -4, -3, -2, -1, 0, 1, 2$$, which is 8 integers (from $$-5$$ to $$2$$ inclusive).

The required probability is $$\frac{8}{36} = \frac{2}{9}$$.

This corresponds to Option 2.